[HEOI 2014]大工程

最新推荐文章于 2020-07-31 23:00:58 发布

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最新推荐文章于 2020-07-31 23:00:58 发布

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原文链接：http://www.cnblogs.com/NaVi-Awson/p/8456238.html

本文介绍了一种解决特定树形结构问题的方法——通过构建虚树并运用树形DP技术来求解关键点间的路径长度和及其最大最小值。文章提供了一个完整的C++代码示例，展示了如何有效地处理大规模数据集上的复杂查询。

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Description

题库链接

给你一个 \(n\) 个节点的树， \(q\) 组询问，每次给出 \(k\) 个关键点，询问这 \(k\) 个关键点两两间路径长度和，长度最值。

\(1\leq n\leq 1000000, 1\leq q\leq 50000,\sum k\leq 2n\)

Solution

建完虚树后，直接简单的树形 \(DP\) ，开四个辅助数组， \(size,sum,maxn,minn\) 表示虚树中以它为根的子树中所有关键点的个数，到它的路径长度和，长度最值。

再一顿乱搞就好了。

Code

//It is made by Awson on 2018.2.21
#include <bits/stdc++.h>
#define LL long long
#define dob complex<double>
#define Abs(a) ((a) < 0 ? (-(a)) : (a))
#define Max(a, b) ((a) > (b) ? (a) : (b))
#define Min(a, b) ((a) < (b) ? (a) : (b))
#define Swap(a, b) ((a) ^= (b), (b) ^= (a), (a) ^= (b))
#define writeln(x) (write(x), putchar('\n'))
#define lowbit(x) ((x)&(-(x)))
using namespace std;
const int N = 1000000;
const int INF = ~0u>>1;
void read(int &x) {
    char ch; bool flag = 0;
    for (ch = getchar(); !isdigit(ch) && ((flag |= (ch == '-')) || 1); ch = getchar());
    for (x = 0; isdigit(ch); x = (x<<1)+(x<<3)+ch-48, ch = getchar());
    x *= 1-2*flag;
}
void print(LL x) {if (x > 9) print(x/10); putchar(x%10+48); }
void write(LL x) {if (x < 0) putchar('-'); print(Abs(x)); }

int n, lim, fa[N+5][25], dep[N+5], dfn[N+5], times, u, v;
LL maxans, minans, sumans, minn[N+5], maxn[N+5], sum[N+5], size[N+5];
int lst[N+5], flag[N+5], k, t, S[N+5], top;
struct graph {
    struct tt {int to, next; }edge[(N<<1)+5];
    int path[N+5], top;
    void add(int u, int v) {edge[++top].to = v, edge[top].next = path[u], path[u] = top; }
    void dfs1(int o, int depth) {
    dep[o] = depth, dfn[o] = ++times;
    for (int i = 1; i <= lim; i++) fa[o][i] = fa[fa[o][i-1]][i-1];
    for (int i = path[o]; i; i = edge[i].next)
        if (dfn[edge[i].to] == 0) fa[edge[i].to][0] = o, dfs1(edge[i].to, depth+1);
    }
    void dfs2(int o) {
    minn[o] = INF, maxn[o] = -INF, sum[o] = 0, size[o] = 0;
    if (flag[o]) minn[o] = maxn[o] = 0, size[o] = 1;
    for (int &i = path[o]; i; i = edge[i].next) {
        dfs2(edge[i].to);
        sumans += size[o]*(sum[edge[i].to]+(dep[edge[i].to]-dep[o])*size[edge[i].to])+sum[o]*size[edge[i].to];
        sum[o] += sum[edge[i].to]+size[edge[i].to]*(dep[edge[i].to]-dep[o]), size[o] += size[edge[i].to];
        minans = Min(minans, minn[o]+minn[edge[i].to]+dep[edge[i].to]-dep[o]), minn[o] = Min(minn[o], minn[edge[i].to]+dep[edge[i].to]-dep[o]);
        maxans = Max(maxans, maxn[o]+maxn[edge[i].to]+dep[edge[i].to]-dep[o]), maxn[o] = Max(maxn[o], maxn[edge[i].to]+dep[edge[i].to]-dep[o]);
    }
    }
}g1, g2;
bool comp(const int &a, const int &b) {return dfn[a] < dfn[b]; }
int get_lca(int u, int v) {
    if (dep[u] < dep[v]) Swap(u, v);
    for (int i = lim; i >= 0; i--) if (dep[fa[u][i]] >= dep[v]) u = fa[u][i];
    if (u == v) return u;
    for (int i = lim; i >= 0; i--) if (fa[u][i] != fa[v][i]) u = fa[u][i], v = fa[v][i];
    return fa[u][0];
}

void work() {
    read(n); lim = log(n)/log(2);
    for (int i = 1; i < n; i++) read(u), read(v), g1.add(u, v), g1.add(v, u);
    g1.dfs1(1, 1); read(t);
    while (t--) {
    read(k); top = g2.top = 0; maxans = -INF, minans = INF, sumans = 0;
    for (int i = 1; i <= k; i++) read(lst[i]), flag[lst[i]] = 1;
    sort(lst+1, lst+1+k, comp); S[++top] = lst[1];
    for (int i = 2; i <= k; i++) {
        int lca = get_lca(lst[i], S[top]);
        while (dfn[lca] < dfn[S[top]]) {
        if (dfn[lca] >= dfn[S[top-1]]) {
            g2.add(lca, S[top]); --top;
            if (S[top] != lca) S[++top] = lca;
            break;
        }
        g2.add(S[top-1], S[top]), --top;
        }
        S[++top] = lst[i];
    }
    while (top > 1) g2.add(S[top-1], S[top]), --top;
    g2.dfs2(S[1]);
    write(sumans), putchar(' '), write(minans), putchar(' '), writeln(maxans);
    for (int i = 1; i <= k; i++) flag[lst[i]] = 0;
    }
}
int main() {
    work(); return 0;
}

转载于:https://www.cnblogs.com/NaVi-Awson/p/8456238.html